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May 21

Linear recurrence

Find a closed form for f(n)=3f(n-1)-2f(n-2), f(0)=2, f(1)=1.

Here's how I solved it, but the equation at the end produces the wrong value for f(2); it should be -1, but the equation gives -7.

f(n)=3f(n-1)-2f(n-2)

x^n=3x^{n-1}-2x^{n-2}

x^2-3x+2=0

x=-1,-2

f(n)=ax^n+by^n

f(n)=a(-1)^n+b(-2)^n

2=a+b

1=-a-2b

a=5, b=-3

f(n)=5(-1)^n-3(-2)^n

--70.129.186.243 (talk) 01:12, 21 May 2010 (UTC)[reply]

That's an interesting way to solve these linear recurrences. Anyway, your problem is in this step

x^2-3x+2=0

x=-1,-2

Those are not the solutions to that quadratic equation. With the correct roots, I think you'll get the final answer with no trouble.98.235.80.144 (talk) 02:15, 21 May 2010 (UTC)[reply]

The OP is using the method described at Recurrence relation#Theorem.—Emil J. 12:46, 21 May 2010 (UTC)[reply]

It looks like you misfactored, 70. How are the solutions negative if the linear term of the quadratic is negative? 76.229.164.175 (talk) 02:23, 21 May 2010 (UTC)[reply]

behaviour of rotating hardboiled egg

Could a kind mathematician (possibly one who has read the Keith Moffat Yutaka Shimomura paper) give an answer at Wikipedia:Reference_desk/Science#Spinning_Tesla_egg_and_stability - is the Moffat/Shinomura paper right? Does friction off centre of rotation cause the egg to try to roll as it rotates - and then trace a path on the surface of the egg, or what?? Is the use of the euler equation idea described there correct.. etc . Thanks. I think the part about the ac motor is understood - it's the behaviour of a spinning egg that confuses.77.86.115.45 (talk) 02:33, 21 May 2010 (UTC)[reply]

sum continous random variable

If you know the pdfs of two continuous random variables, what is the pdf of the sum of those two variables? --115.178.29.142 (talk) 03:31, 21 May 2010 (UTC)[reply]

It's the convolution of the two pdfs. I'm pretty sure we have an article about this; I'll see if I can find it. Michael Hardy (talk) 03:38, 21 May 2010 (UTC)[reply]

The article section Probability_distribution#Some_properties confirms what Michael said. But instead of the pdf you might use the cumulant generating function (cgf), because the cgf of a sum is the sum of the cgfs. Follow the simplicity! Bo Jacoby (talk) 09:54, 21 May 2010 (UTC).[reply]

That depends on what you're trying to do. Michael Hardy (talk) 20:18, 21 May 2010 (UTC)[reply]

Finding a geodesic

How do you find the formula for a geodesic using only the metric tensor? I've asked this before, but by the time I figured out what to ask, I think it was too late to get much of a response.

I know it can't generally be done with elementary functions, but I know it at least could be done with a Taylor series. Can someone tell me how, or where to find out how?

Should I be using something other than the metric tensor? If so, what?

My math is up to matrices. I've learned about I haven't taken anything about Riemannian geometry. I'm willing to learn more math if necessary, but I need a good place to do so. Wikibooks doesn't have a book on Riemannian geometry. — DanielLC 05:12, 21 May 2010 (UTC)[reply]

The topic you're asking about is part of differential geometry. We have a good article differential geometry of surfaces which doesn't say much about metric tensors. You probably have to read an actual textbook or take a class. I remember a book by doCarmo ("Differential Geometry of Curves and Surfaces") being pretty accessible though I didn't read it. Maybe your library has it or can get hold of it. 69.228.170.24 (talk) 08:01, 21 May 2010 (UTC)[reply]

The equations for the geodesic are most easily obtained by writing down the Euler-Lagrange equations for the Lagrangian:

$L={\frac {1}{2}}g_{\mu \nu }{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}$

Count Iblis (talk) 15:05, 21 May 2010 (UTC)[reply]

That equation doesn't seem to be the Euler–Lagrange equation, and I don't know what you mean by Lagrangian. Also, I don't know what that stuff means. Does ${\frac {dx^{\mu }}{d\tau }}$ mean the partial derivative of mu in terms of tau?

I guess I just want it in terms of an intrinsic measure of curvature. — DanielLC 18:42, 21 May 2010 (UTC)[reply]

That expression is written using Einstein notation.

{\frac {dx^{\mu }}{d\tau }}

means the derivative of the μth coordinate of the vector x with respect to τ. --Tango (talk) 00:13, 22 May 2010 (UTC)[reply]

I would recommend Serge Lang's Fundamentals of differential geometry, if you wish to either specialize in the subject, or pursue an area which requires a strong background in differential geometry. The book may be slightly advanced, but even if you are not ready for it now, it will probably be extremely useful later. (The link for the "Google books" preview of the book is [1].) PS T 10:45, 22 May 2010 (UTC)[reply]

Newton's division

I have stumbled across Newton's division article. But wouldn't it be more accurate and quick to simply flip a divisor over than to do stuff like it describes? When would it be a good idea to use it, and not to use it? 99.154.83.190 (talk) 21:00, 21 May 2010 (UTC)[reply]

We don't have an article titled Newton's division. What article are you talking about? Algebraist 21:17, 21 May 2010 (UTC)[reply]

(edit conflict)There's no article called Newton's division. Are you referring to Division (digital)#Newton–Raphson_division ? This works by first finding the reciprocal of the divisor, i.e. flipping the divisor over, then multiplying the numerator by that reciprocal. I don't know if it's used by any current computers, but it certainly used to be; see e.g. doi:10.1007/BF02307376, which mentions its use by the Astronautics ZS-1, the Tera Computer, the Intel i860, and the IBM RS/6000. That section of the article could do with some references. --Qwfp (talk) 21:25, 21 May 2010 (UTC)[reply]